y=2sin(3x-2π) 1的单调区间

(1)振幅为1,周期为2π/2=π,初相位为π/3(2)可由y=sinx先横坐标变为原来的2倍,然后左移π/6

sinx的周期是2pai,sin3x的周期是三分之二pai,sin5x的周期是五分之二pai取其最小公倍数,则y的周期是2pai.

x=0:0.01:1;y=0;fori=1:20y=y+sin(i*x);endplot(y);

这个是和差化积公式如没学过,可以这样sin(x+y)-sinx=sin[(x+1/2y)+1/2y]-sin[(x+1/2y)-1/2y]=sin(x+1/2y)cos(1/2y)+cos(x+1/2

你用积化和差公式一套,然后就能看出它的最小正周期来的.应该是1pi

y=sin(x+π/3）sin(x+π/2)=(sinx+√3cosx)cosx/2=[sinxcosx+√3(cosx)^2]/2=[sin2x/2+√3(cos2x+1)/2]/2=(sin2x+

x=-π/6时,y=0所以,关于点(-π/6,0)对称选B

sin(x+y)=sinxcosy+cosxsiny=1/2sin(x-y)=sinxconsy-cosxsiny=1/3sinxcosy=5/12,cosxsiny=1/12tanx/tany=si

把两个三角函数展开,得y=3/2sinx-√3/2cosx合并成：y=√3sin(x-π/6)单调区间是（-π/3,2π/3）增（2π/3,5π/3）减其中都要加上2kπ,我就不写了

原式=(sinx/2+根号3cosx/2)cosx=sinxcosx/2+根号3cos^2x/2=sin2x/4+根号3cos2x/4+根号3/4=2sin(2x+π/3)+根号3/4T=2π/W=π

积化和差公式sinαsinβ=-1/2[cos(α+β)-cos(α-β)]令sinα＝sin(x+π/3）sinβ＝sin（x+π/2）话说到这一步了应该可以自己解决了吧这个高考不要求的吧!积化和差

y=2sin[(1/2)x-π/3]+1(1)最小正周期:T=2π/(1/2)=4π.（2）单调性:由2kπ-π/2≦(1/2)x-π/3≦2kπ+π/2,得2kπ-π/6≦(1/2)x≦2kπ+5π

x+3>0,且ln(x+3)≠0得：x>-3且x≠-2所以,定义域为(-3,-2)U(-2,+∞)

sin^2x+cos^2y=1/2∴sin^2x=1/2-cos^2y3sin^2x+sin^2y=3（1/2-cos^2y）+sin^2y=1.5-3cos^2y）+sin^2y又有sin^2y+c

y=sin(x+π/3)sin(x+π/2)=sin(x+π/3)cosx=(sinxcosπ/3+cosxsinπ/3)cosx=1/2sinxcosx+√3/2cos^2(x)[cos^2(x)指

Sinx-siny=2/3cosx-cosy=1/2分别平方得(Sinx-siny)^2=(2/3)^2(cosx-cosy)^2=(1/2)^2展开相加得-2cos(x-y)+2=4/9+1/4-2

因为cos(2x)=1-2sin^2(x),所以sin^2(x)=[1-cos(2x)]/2.y=1/2sin(2x)+sin^2(x)=1/2sin(2x)+[1-cos(2x)]/2=1/2*si

y=1/2sin(π/4-2x/3)=-1/2sin(2x/3-π/4)不考虑周期时,根据正弦在[-π/2,π/2]递增,在[π/2,3π/2]递减结合本题前面有一个负号,则增减相反得单调递增区间（[

y=2sin(2x+π/3)+sin(2x-π/3)=2(sin2xcos派/3+cos2xsin派/3)+sin2xcos派/3-cos2xsin派/3=3sin2xcos派/3+cos2xsin派